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Section 11-1 Tangent Lines SPI 32B: Identify chords of circles given a diagram SPI 33A: Solve problems involving the properties of arcs, tangents, chords.

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Presentation on theme: "Section 11-1 Tangent Lines SPI 32B: Identify chords of circles given a diagram SPI 33A: Solve problems involving the properties of arcs, tangents, chords."— Presentation transcript:

1 Section 11-1 Tangent Lines SPI 32B: Identify chords of circles given a diagram SPI 33A: Solve problems involving the properties of arcs, tangents, chords Objectives: Use relationships between a radius and a tangent Tangent to a Circle Line in the plane of a circle that intersects the circle in exactly 1 point (Line AB is tangent to the circle) A B Point of Tangency Point where circle and a tangent intersect (B is the point of Tangency)

2 Relate Tangent and Radius of a Circle

3 BA is tangent to C at point A. Find the value of x x = 180 Substitute x = 180Simplify. x = 68 Solve. m A + m B + m C = 180 Triangle Angle-Sum Theorem Because BA is tangent to C, A must be a right angle. Use the Triangle Angle-Sum Theorem to find x.. Tangent Lines

4 Because opposite sides of a rectangle have the same measure, DW = 3 cm and OD = 15 cm. Because OZ is a radius of O, OZ = 3 cm.. A belt fits tightly around two circular pulleys, as shown below. Find the distance between the centers of the pulleys. Round your answer to the nearest tenth. Draw OP. Then draw OD parallel to ZW to form rectangle ODWZ, as shown below. Real World and Tangent Lines

5 OD 2 + PD 2 = OP 2 Pythagorean Theorem = OP 2 Substitute. 241 = OP 2 Simplify. The distance between the centers of the pulleys is about 15.5 cm. OP Use a calculator to find the square root. Because the radius of P is 7 cm, PD = 7 – 3 = 4 cm.. Because ODP is the supplement of a right angle, ODP is also a right angle, and OPD is a right triangle. (continued)

6 Finding a Tangent

7 144 = 194Simplify. / O has radius 5. Point P is outside O such that PO = 12, and point A is on O such that PA = 13. Is PA tangent to O at A? Explain Substitute. Because PO 2 = PA 2 + OA 2, PA is not tangent to O at A. /. PO 2 PA 2 + OA 2 Is OAP a right triangle? Draw the situation described in the problem. For PA to be tangent to O at A, A must be a right angle, OAP must be a right triangle, and PO 2 = PA 2 + OA 2.. Tangent Lines

8 Using Multiple Tangents When a circle is inscribed in a triangle, the triangle is circumscribed about the circle. What is the relationship between each side of the triangle and the circle? Each segment is tangent to the circle, meaning each line is perpendicular to the radius forming a right angle.

9 Using Tangents to Solve Problems

10 QS and QT are tangent to O at points S and T, respectively. Give a convincing argument why the diagonals of quadrilateral QSOT are perpendicular.. Theorem 11-3 states that two segments tangent to a circle from a point outside the circle are congruent. OS = OT because all radii of a circle are congruent. Two pairs of adjacent sides are congruent. Quadrilateral QSOT is a kite if no opposite sides are congruent or a rhombus if all sides are congruent. By theorems in Lessons 6-4 and 6-5, both the diagonals of a rhombus and the diagonals of a kite are perpendicular. Because QS and QT are tangent to O, QS QT, so QS = QT.. Using Theorem

11 p = XY + YZ + ZW + WX Definition of perimeter p = XR + RY + YS + SZ + ZT + TW + WU + UX Segment Addition Postulate = Substitute. = 64 Simplify. The perimeter is 64 ft. XU = XR = 11 ft YS = YR = 8 ft ZS = ZT = 6 ft WU = WT = 7 ft By Theorem 11-3, two segments tangent to a circle from a point outside the circle are congruent. C is inscribed in quadrilateral XYZW. Find the perimeter of XYZW.. Tangent Lines


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